Question

A small box of mass m is placed on a top of a larger box of mass 2m as shown in the diagram. When a force F is applied to the large box, both boxes accelerate to the right with the same acceleration. If the coefficient of friction between all surfaces is $\mu$ what would be the force accelerating the smaller mass?

• A. $\frac{F}{3}-mg\mu$
• B. $F-3mg\mu$
• C. $F-mg\mu$
• D. $\frac{F-mg\mu }{3}$

Answer 1

Answer: (A)

$\frac{F}{3}-mg\mu$

both boxes accelerates with same acceleration then we assume it as a body of mass '3m' and moving with acceleration 'a' .

friction at the ground surface is

$f=\mu NN=3mg=\mu \left(3mg\right)=3\mu mg$

then acceleration is

$a=\frac{F-3\mu mg}{3m}=\frac{F}{3m}-\mu g$

if mass m is moving with same acceleration 'a'

then force applied on it is

$F^{\prime }=ma=\frac{F}{3}-\mu mg$

Option A is correct